A body of mass `m_(1)` collides elastically with a stationary body of mass `m_(2)` and returns along the same line with one fourth of its initial speed, then `m_(1)//m_(2)=`

Options

A. 1
B. 2
C. `0.66`
D. `0.33`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
We have two bodies:
- Mass \( m_1 \) is moving with an initial velocity \( v \).
- Mass \( m_2 \) is stationary (initial velocity = 0).

After the elastic collision, mass \( m_1 \) returns with a velocity of \( -\frac{v}{4} \) (negative because it is in the opposite direction).

### Step 2: Apply the conservation of momentum
The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

Before the collision:
- Momentum of \( m_1 = m_1 \cdot v \)
- Momentum of \( m_2 = m_2 \cdot 0 = 0 \)

Total initial momentum:
\[ p_{\text{initial}} = m_1 \cdot v + 0 = m_1 v \]

After the collision:
- Momentum of \( m_1 = m_1 \cdot \left(-\frac{v}{4}\right) \)
- Momentum of \( m_2 = m_2 \cdot v_2 \) (where \( v_2 \) is the velocity of \( m_2 \) after the collision)

Total final momentum:
\[ p_{\text{final}} = -m_1 \cdot \frac{v}{4} + m_2 \cdot v_2 \]

Setting the initial momentum equal to the final momentum:
\[ m_1 v = -m_1 \cdot \frac{v}{4} + m_2 \cdot v_2 \tag{1} \]

### Step 3: Apply the conservation of kinetic energy
Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

Before the collision:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 v^2 + 0 = \frac{1}{2} m_1 v^2 \]

After the collision:
\[ KE_{\text{final}} = \frac{1}{2} m_1 \left(-\frac{v}{4}\right)^2 + \frac{1}{2} m_2 v_2^2 \]
\[ = \frac{1}{2} m_1 \cdot \frac{v^2}{16} + \frac{1}{2} m_2 v_2^2 \]

Setting the initial kinetic energy equal to the final kinetic energy:
\[ \frac{1}{2} m_1 v^2 = \frac{1}{2} m_1 \cdot \frac{v^2}{16} + \frac{1}{2} m_2 v_2^2 \tag{2} \]

### Step 4: Solve the equations
From equation (1):
\[ m_1 v + m_1 \cdot \frac{v}{4} = m_2 v_2 \]
\[ m_1 \left(v + \frac{v}{4}\right) = m_2 v_2 \]
\[ m_1 \cdot \frac{5v}{4} = m_2 v_2 \tag{3} \]

From equation (2):
\[ m_1 v^2 = m_1 \cdot \frac{v^2}{16} + m_2 v_2^2 \]
\[ m_1 v^2 - m_1 \cdot \frac{v^2}{16} = m_2 v_2^2 \]
\[ m_1 \left(v^2 - \frac{v^2}{16}\right) = m_2 v_2^2 \]
\[ m_1 \cdot \frac{15v^2}{16} = m_2 v_2^2 \tag{4} \]

### Step 5: Substitute \( v_2 \) from equation (3) into equation (4)
From equation (3):
\[ v_2 = \frac{5m_1 v}{4m_2} \]

Substituting \( v_2 \) into equation (4):
\[ m_1 \cdot \frac{15v^2}{16} = m_2 \left(\frac{5m_1 v}{4m_2}\right)^2 \]
\[ m_1 \cdot \frac{15v^2}{16} = m_2 \cdot \frac{25m_1^2 v^2}{16m_2^2} \]
\[ m_1 \cdot 15 = \frac{25m_1^2}{m_2} \]

### Step 6: Solve for \( \frac{m_1}{m_2} \)
Rearranging gives:
\[ 15m_2 = 25m_1 \]
\[ \frac{m_1}{m_2} = \frac{15}{25} = \frac{3}{5} \]

### Final Answer
Thus, the ratio \( \frac{m_1}{m_2} \) is \( \frac{3}{5} \).

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